Toronto Math Forum

MAT244--2019F => MAT244--Test & Quizzes => Quiz-4 => Topic started by: Ruojing Chen on October 18, 2019, 02:02:34 PM

Title: tut 0202 quiz4
Post by: Ruojing Chen on October 18, 2019, 02:02:34 PM
Find the general solution for equation y''-2y'+6y=0

$$ set \ r^2-2r+6=0$$

$$r=\frac{2\pm\sqrt{(-2)^2-4*6}}{2}$$

$$r=\frac{2\pm2i\sqrt{5}}{2}$$

$$r=1\pm\sqrt{5}i$$

$$y_c(t)=c_1e^tCos(\sqrt{5}t)+c_2e^tSin(\sqrt{5}t)$$